• 358,603

# Posts Tagged ‘four momentum’

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## Relativistic generalization of statistical mechanics

Posted by peeterjoot on March 22, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation

I was wondering how to generalize the arguments of [1] to relativistic systems. Here’s a bit of blundering through the non-relativistic arguments of that text, tweaking them slightly.

I’m sure this has all been done before, but was a useful exercise to understand the non-relativistic arguments of Pathria better.

# Generalizing from energy to four momentum

Generalizing the arguments of section 1.1.

Instead of considering that the total energy of the system is fixed, it makes sense that we’d have to instead consider the total four-momentum of the system fixed, so if we have $N$ particles, we have a total four momentum

\begin{aligned}P = \sum_i n_i P_i = \sum n_i \left( \epsilon_i/c, \mathbf{p}_i \right),\end{aligned} \hspace{\stretch{1}}(1.2.1)

where $n_i$ is the total number of particles with four momentum $P_i$. We can probably expect that the $n_i$‘s in this relativistic system will be smaller than those in a non-relativistic system since we have many more states when considering that we can have both specific energies and specific momentum, and the combinatorics of those extra degrees of freedom. However, we’ll still have

\begin{aligned}N = \sum_i n_i.\end{aligned} \hspace{\stretch{1}}(1.2.2)

Only given a specific observer frame can these these four-momentum components $\left( \epsilon_i/c, \mathbf{p}_i \right)$ be expressed explicitly, as in

\begin{aligned}\epsilon_i = \gamma_i m_i c^2\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{p}_i = \gamma_i m \mathbf{v}_i\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\begin{aligned}\gamma_i = \frac{1}{{\sqrt{1 - \mathbf{v}_i^2/c^2}}},\end{aligned} \hspace{\stretch{1}}(1.0.3c)

where $\mathbf{v}_i$ is the velocity of the particle in that observer frame.

# Generalizing the number if microstates, and notion of thermodynamic equilibrium

Generalizing the arguments of section 1.2.

We can still count the number of all possible microstates, but that number, denoted $\Omega(N, V, E)$, for a given total energy needs to be parameterized differently. First off, any given volume is observer dependent, so we likely need to map

\begin{aligned}V \rightarrow \int d^4 x = \int dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3.\end{aligned} \hspace{\stretch{1}}(1.0.4)

Let’s still call this $V$, but know that we mean this to be four volume element, bounded in both space and time, referred to a fixed observer’s frame. So, lets write the total number of microstates as

\begin{aligned}\Omega(N, V, P) = \Omega \left( N, \int d^4 x, E/c, P^1, P^2, P^3 \right),\end{aligned} \hspace{\stretch{1}}(1.0.5)

where $P = ( E/c, \mathbf{P} )$ is the total four momentum of the system. If we have a system subdivided into to two systems in contact as in fig. 1.1, where the two systems have total four momentum $P_1$ and $P_2$ respectively.

Fig 1.1: Two physical systems in thermal contact

In the text the total energy of both systems was written

\begin{aligned}E^{(0)} = E_1 + E_2,\end{aligned} \hspace{\stretch{1}}(1.0.6)

so we’ll write

\begin{aligned}{P^{(0)}}^\mu = P_1^\mu + P_2^\mu = \text{constant},\end{aligned} \hspace{\stretch{1}}(1.0.7)

so that the total number of microstates of the combined system is now

\begin{aligned}\Omega^{(0)}(P_1, P_2) = \Omega_1(P_1) \Omega_2(P_2).\end{aligned} \hspace{\stretch{1}}(1.0.8)

As before, if $\bar{{P}}^\mu_i$ denotes an equilibrium value of $P_i^\mu$, then maximizing eq. 1.0.8 requires all the derivatives (no sum over $\mu$ here)

\begin{aligned}\left({\partial {\Omega_1(P_1)}}/{\partial {P^\mu_1}}\right)_{{P_1 = \bar{{P_1}}}}\Omega_2(\bar{{P}}_2)+\Omega_1(\bar{{P}}_1)\left({\partial {\Omega_2(P_2)}}/{\partial {P^\mu}}\right)_{{P_2 = \bar{{P_2}}}}\times\frac{\partial {P_2^\mu}}{\partial {P_1^\mu}}= 0.\end{aligned} \hspace{\stretch{1}}(1.0.9)

With each of the components of the total four-momentum $P^\mu_1 + P^\mu_2$ separately constant, we have ${\partial {P_2^\mu}}/{\partial {P_1^\mu}} = -1$, so that we have

\begin{aligned}\left({\partial {\ln \Omega_1(P_1)}}/{\partial {P^\mu_1}}\right)_{{P_1 = \bar{{P_1}}}}=\left({\partial {\ln \Omega_2(P_2)}}/{\partial {P^\mu}}\right)_{{P_2 = \bar{{P_2}}}},\end{aligned} \hspace{\stretch{1}}(1.0.10)

as before. However, we now have one such identity for each component of the total four momentum $P$ which has been held constant. Let’s now define

\begin{aligned}\beta_\mu \equiv \left({\partial {\ln \Omega(N, V, P)}}/{\partial {P^\mu}}\right)_{{N, V, P = \bar{{P}}}},\end{aligned} \hspace{\stretch{1}}(1.0.11)

Our old scalar temperature is then

\begin{aligned}\beta_0 = c \left({\partial {\ln \Omega(N, V, P)}}/{\partial {E}}\right)_{{N, V, P = \bar{{P}}}} = c \beta = \frac{c}{k_{\mathrm{B}} T},\end{aligned} \hspace{\stretch{1}}(1.0.12)

but now we have three additional such constants to figure out what to do with. A first start would be figuring out how the Boltzmann probabilities should be generalized.

Equilibrium between a system and a heat reservoir

Generalizing the arguments of section 3.1.

As in the text, let’s consider a very large heat reservoir $A'$ and a subsystem $A$ as in fig. 1.2 that has come to a state of mutual equilibrium. This likely needs to be defined as a state in which the four vector $\beta_\mu$ is common, as opposed to just $\beta_0$ the temperature field being common.

Fig 1.2: A system A immersed in heat reservoir A’

If the four momentum of the heat reservoir is $P_r'$ with $P_r$ for the subsystem, and

\begin{aligned}P_r + P_r' = P^{(0)} = \text{constant}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

Writing

\begin{aligned}\Omega'({P^\mu_r}') = \Omega'(P^{(0)} - {P^\mu_r}) \propto P_r,\end{aligned} \hspace{\stretch{1}}(1.0.14)

for the number of microstates in the reservoir, so that a Taylor expansion of the logarithm around $P_r' = P^{(0)}$ (with sums implied) is

\begin{aligned}\ln \Omega'({P^\mu_r}') = \ln \Omega'({P^{(0)}}) +\left({\partial {\ln \Omega'}}/{\partial {{P^\mu}'}}\right)_{{P' = P^{(0)}}} \left( P^{(0)} - P^\mu \right)\approx\text{constant} - \beta_\mu' P^\mu.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Here we’ve inserted the definition of $\beta^\mu$ from eq. 1.0.11, so that at equilibrium, with $\beta_\mu' = \beta_\mu$, we obtain

\begin{aligned}\Omega'({P^\mu_r}') = \exp\left( - \beta_\mu P^\mu \right)=\exp\left( - \beta E \right)\exp\left( - \beta_1 P^1 \right)\exp\left( - \beta_2 P^3 \right)\exp\left( - \beta_3 P^3 \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

# Next steps

This looks consistent with the outline provided in http://physics.stackexchange.com/a/4950/3621 by Lubos to the stackexchange “is there a relativistic quantum thermodynamics” question. I’m sure it wouldn’t be too hard to find references that explore this, as well as explain why non-relativistic stat mech can be used for photon problems. Further exploration of this should wait until after the studies for this course are done.

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## Gauge transformation of the Dirac equation.

Posted by peeterjoot on August 21, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

In [1] the gauge transformation of the Dirac equation is covered, producing the non-relativistic equation with the correct spin interaction. There are unfortunately some sign errors, some of which self correct, and some of which don’t impact the end result, but are slightly confusing. There are also some omitted details. I’ll attempt to work through the same calculation with all the signs in the right places and also fill in some of the details I found myself wanting.

# A step back. On the gauge transformation.

The gauge transformations utilized are given as

\begin{aligned}\mathcal{E} &\rightarrow \mathcal{E} - e \phi \\ \mathbf{p} &\rightarrow \mathbf{p} - e \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.1)

Let’s start off by reminding ourself where these come from. As outlined in section 12.9 in [2] (with some details pondered in [3]), our relativistic Lagrangian is

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{ 1 - \frac{\mathbf{u}}{c^2}} + \frac{e}{c} \mathbf{u} \cdot \mathbf{A} - e \phi.\end{aligned} \hspace{\stretch{1}}(2.3)

The conjugate momentum is

\begin{aligned}\mathbf{P} = \mathbf{e}^i \frac{\partial {\mathcal{L}}}{\partial {u^i}} = \frac{m \mathbf{u}}{\sqrt{1 - \mathbf{u}^2/c^2}} + \frac{e}{c} \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.4)

or

\begin{aligned}\mathbf{P} = \mathbf{p} + \frac{e}{c} \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.5)

The Hamiltonian, which must be expressed in terms of this conjugate momentum $\mathbf{P}$, is found to be

\begin{aligned}\mathcal{E} = \sqrt{ (c \mathbf{P} - e \mathbf{A})^2 + m^2 c^4 } + e \phi.\end{aligned} \hspace{\stretch{1}}(2.6)

With the free particle Lagrangian

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{ 1 - \frac{\mathbf{u}}{c^2}} ,\end{aligned} \hspace{\stretch{1}}(2.7)

our conjugate momentum is

\begin{aligned}\mathbf{P} = \frac{m \mathbf{u}}{\sqrt{ 1 - \mathbf{u}^2/c^2} }.\end{aligned} \hspace{\stretch{1}}(2.8)

For this we find that our Hamiltonian $\mathcal{E} = \mathbf{P} \cdot \mathbf{u} - \mathcal{L}$ is

\begin{aligned}\mathcal{E} = \frac{m c^2}{\sqrt{1 - \mathbf{u}^2/c^2}},\end{aligned} \hspace{\stretch{1}}(2.9)

but this has to be expressed in terms of $\mathbf{P}$. Having found the form of the Hamiltonian for the interaction case, it is easily verified that 2.6 contains the required form once the interaction fields $(\phi, \mathbf{A})$ are zeroed

\begin{aligned}\mathcal{E} = \sqrt{ (c \mathbf{P})^2 + m^2 c^4 }.\end{aligned} \hspace{\stretch{1}}(2.10)

Considering the interaction case, Jackson points out that the energy and momentum terms can be combined as a four momentum

\begin{aligned}p^a = \left( \frac{1}{{c}}(\mathcal{E} - e \phi), \mathbf{P} - \frac{e}{c}\mathbf{A} \right),\end{aligned} \hspace{\stretch{1}}(2.11)

so that the re-arranged and squared Hamiltonian takes the form

\begin{aligned}p^a p_a = (m c)^2.\end{aligned} \hspace{\stretch{1}}(2.12)

From this we see that for the Lorentz force, the interaction can be found, starting with the free particle Hamiltonian 2.6, making the transformation

\begin{aligned}\mathcal{E} &\rightarrow \mathcal{E} - e\phi \\ \mathbf{P} &\rightarrow \mathbf{P} - \frac{e}{c}\mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.13)

or in covariant form

\begin{aligned}p^\mu \rightarrow p^\mu - \frac{e}{c}A^\mu.\end{aligned} \hspace{\stretch{1}}(2.15)

# On the gauge transformation of the Dirac equation.

The task at hand now is to make the transformations of 2.13, applied to the Dirac equation

\begin{aligned}{p} = \gamma_\mu p^\mu = m c.\end{aligned} \hspace{\stretch{1}}(3.16)

The first observation to make is that we appear to have different units in the Desai text. Let’s continue using the units from Jackson, and translate them later if inclined.

Right multiplication of 3.16 by $\gamma_0$ gives us

\begin{aligned}0 &= \gamma_0 ({p} - m c) \\ &= \gamma_0 \gamma_\mu \left( p^\mu - \frac{e}{c} A^\mu \right)- \gamma_0 m c\\ &=\gamma_0 \gamma_0 \left(\frac{\mathcal{E}}{c} - \frac{e}{c} \phi \right)+\gamma_0 \gamma_a \left(p^a - \frac{e}{c} A^a \right)- \gamma_0 m c \\ &=\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \\ \end{aligned}

With the minor notational freedom of using $\gamma_0$ instead of $\gamma_4$, this is our starting point in the Desai text, and we can now left multiply by

\begin{aligned}({p} + m c) \gamma_0 =\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)+\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)+ \gamma_0 m c.\end{aligned} \hspace{\stretch{1}}(3.17)

The motivation for this appears to be that this product of conjugate like quantities

\begin{aligned}\begin{aligned}0 &= ({p} + m c) \gamma_0 \gamma_0 ({p} - m c) \\ &=({p} + m c) ({p} - m c) \\ &= \frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2 -\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2 - (m c)^2 + \cdots,\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.18)

produces the the Klein-Gordon equation, plus some cross terms to be determined. Those cross terms are the important bits since they contain the spin interaction, even in the non-relativistic limit.

Let’s do the expansion.

\begin{aligned}0&= ({p} + m c) \gamma_0 \gamma_0 ({p} - m c) u \\ &=\left(\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)+\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)+ \gamma_0 m c\right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &=\frac{1}{{c}} \left( \mathcal{E} - e \phi \right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &\qquad +\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &\qquad + \gamma_0 m c\left(\frac{1}{{c}} \left( \mathcal{E}- e \phi \right)-\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)- \gamma_0 m c \right) u \\ &=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left( \boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right) \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}} \left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u- m c\left\{{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{ \gamma_0}\right\} u \\ &\qquad + {\gamma_0 m\left(\mathcal{E} - e \phi\right) u}- {\gamma_0 m\left(\mathcal{E} - e \phi\right) u}\\ \end{aligned}

Since $\gamma_0$ anticommutes with any $\boldsymbol{\alpha} \cdot \mathbf{x}$, even when $\mathbf{x}$ contains operators, the anticommutator term is killed.

While done in the text, lets also do the $\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)$ square for completeness. Because this is an operator, we need to treat this as

\begin{aligned}\left( \boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right) \right)^2 u&=\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)\boldsymbol{\alpha} \cdot \left(\mathbf{P} u - \frac{e}{c} \mathbf{A} u \right),\end{aligned}

so want to treat the two vectors as independent, say $(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})$. That is

\begin{aligned}(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})&=\begin{bmatrix}0 & \boldsymbol{\sigma} \cdot \mathbf{a} \\ \boldsymbol{\sigma} \cdot \mathbf{a} & 0\end{bmatrix}\begin{bmatrix}0 & \boldsymbol{\sigma} \cdot \mathbf{b} \\ \boldsymbol{\sigma} \cdot \mathbf{b} & 0\end{bmatrix} \\ &=\begin{bmatrix}(\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b}) & 0 \\ 0 & (\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b}) & 0 \\ \end{bmatrix} \\ \end{aligned}

The diagonal elements can be expanded by coordinates

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a}) (\boldsymbol{\sigma} \cdot \mathbf{b})&=\sum_{m,n} \sigma^m a^m \sigma^n b^n \\ &=\sum_m a^m b^m+\sum_{m\ne n} \sigma^m \sigma^n a^m b^m \\ &=\mathbf{a} \cdot \mathbf{b}+i \sum_{m\ne n} \sigma^o \epsilon^{m n o} a^m b^m \\ &=\mathbf{a} \cdot \mathbf{b}+i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b}),\end{aligned}

for

\begin{aligned}(\boldsymbol{\alpha} \cdot \mathbf{a})(\boldsymbol{\alpha} \cdot \mathbf{b})=\begin{bmatrix}\mathbf{a} \cdot \mathbf{b} + i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b}) & 0 \\ 0 & \mathbf{a} \cdot \mathbf{b} + i \boldsymbol{\sigma} \cdot (\mathbf{a} \times \mathbf{b})\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

Plugging this back in, we now have an extra term in the expansion

\begin{aligned}0&=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}} \left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u\\ &\qquad- i \boldsymbol{\sigma}' \cdot\left(\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) \times \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right)\right) u\end{aligned}

Here $\boldsymbol{\sigma}'$ was defined as the direct product of the two by two identity with the abstract matrix $\boldsymbol{\sigma}$ as follows

\begin{aligned}\boldsymbol{\sigma}' =\begin{bmatrix}\boldsymbol{\sigma} & 0 \\ 0 & \boldsymbol{\sigma}\end{bmatrix}= I \otimes \boldsymbol{\sigma}\end{aligned} \hspace{\stretch{1}}(3.20)

Like the $\mathbf{L} \times \mathbf{L}$ angular momentum operator cross products this one wasn’t zero. Expanding it yields

\begin{aligned}\left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) \times \left( \mathbf{P} - \frac{e}{c} \mathbf{A} \right) u&=\mathbf{P} \times \mathbf{P} u+ \frac{e^2}{c^2} \mathbf{A} \times \mathbf{A} u- \frac{e}{c} \left( \mathbf{A} \times \mathbf{P} + \mathbf{P} \times \mathbf{A} \right) u \\ &=- \frac{e}{c} \left( \mathbf{A} \times (\mathbf{P} u) + (\mathbf{P} u) \times \mathbf{A} + u (\mathbf{P} \times \mathbf{A}) \right) \\ &=- \frac{e}{c} (-i \hbar \boldsymbol{\nabla} \times \mathbf{A}) u \\ &=\frac{i e \hbar}{c} \mathbf{H} u\end{aligned}

Plugging in again we are getting closer, and now have the magnetic field cross term

\begin{aligned}0&=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2\right) u\\ &\qquad + \frac{1}{{c}}\left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u\\ &\qquad+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H} u.\end{aligned}

All that remains is evaluation of the commutator term, which should yield the electric field interaction. That commutator is

\begin{aligned}\left[{\boldsymbol{\alpha} \cdot \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)},{\mathcal{E} - e \phi}\right] u&={\boldsymbol{\alpha} \cdot \mathbf{P} \mathcal{E} u}- e \boldsymbol{\alpha} \cdot \mathbf{P} \phi u- \frac{e}{c} \boldsymbol{\alpha} \cdot \mathbf{A} \mathcal{E} u+ {\frac{e^2}{c} \boldsymbol{\alpha} \cdot \mathbf{A} \phi u} \\ &- {\mathcal{E} \boldsymbol{\alpha} \cdot \mathbf{P} u}+ e \phi \boldsymbol{\alpha} \cdot \mathbf{P} u+ \frac{e}{c} \mathcal{E} \boldsymbol{\alpha} \cdot \mathbf{A} u- {\frac{e^2}{c} \phi \boldsymbol{\alpha} \cdot \mathbf{A} u} \\ &=\boldsymbol{\alpha} \cdot \left( - e \mathbf{P} \phi+ \frac{e}{c} \mathcal{E} \right) u \\ &=e i \hbar \boldsymbol{\alpha} \cdot \left( \boldsymbol{\nabla} \phi+ \frac{1}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \right) u \\ &=- e i \hbar \boldsymbol{\alpha} \cdot \mathbf{E} u\end{aligned}

That was the last bit required to fully expand the space time split of our squared momentum equations. We have

\begin{aligned}0=({p} + mc)({p} - mc) u=\left(\frac{1}{{c^2}} \left( \mathcal{E} - e \phi \right)^2- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- (mc)^2- \frac{i e \hbar}{c} \boldsymbol{\alpha} \cdot \mathbf{E}+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H}\right) u\end{aligned} \hspace{\stretch{1}}(3.21)

This is the end result of the reduction of the spacetime split gauge transformed Dirac equation. The next step is to obtain the non-relativistic Hamiltonian operator equation (linear in the time derivative operator and quadratic in spacial partials) that has both the electric field and magnetic field terms that we desire to accurately describe spin (actually we need only the magnetic interaction term for non-relativistic spin, but we’ll see that soon).

To obtain the first order time derivatives we can consider an approximation to the $(\mathcal{E} - e \phi)^2$ terms. We can get that by considering the difference of squares factorization

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi - m c^2) ( \mathcal{E} - e \phi + m c^2) u&=\frac{1}{{c^2}} \left(( \mathcal{E} - e \phi )^2 u - (m c^2)^2 u- {m c^2 \mathcal{E} u}+ {\mathcal{E} m c^2 u} \right) \\ &=\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\end{aligned}

In the text, this is factored, instead of the factorization verified. I wanted to be careful to ensure that the operators did not have any effect. They don’t, which is clear in retrospect since the $\mathcal{E}$ operator and the scalar $mc$ necessarily commute. With this factorization, some relativistic approximations are possible. Considering the free particle energy, we can separate out the rest energy from the kinetic (which is perversely designated with subscript $T$ for some reason in the text (and others))

\begin{aligned}\mathcal{E}&= \gamma m c^2 \\ &= m c^2 \left( 1 + \frac{1}{{2}} \left(\frac{\mathbf{v}}{c}\right)^2 + \cdots \right) \\ &= m c^2 + \frac{1}{{2}} m \mathbf{v}^2 + \cdots \\ &\equiv m c^2 + \mathcal{E}_{T}\end{aligned}

With this definition, the energy minus mass term in terms of kinetic energy (that we also had in the Klein-Gordon equation) takes the form

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u=\frac{1}{{c^2}} ( \mathcal{E}_{T} - e \phi ) ( \mathcal{E} - e \phi + m c^2) u\end{aligned} \hspace{\stretch{1}}(3.22)

In the second factor, to get a non-relativistic approximation of $\mathcal{E} - e \phi$, the text states without motivation that $e \phi$ will be considered small compared to $m c^2$. We can make some sense of this by considering the classical Hamiltonian for a particle in a field

\begin{aligned}\mathcal{E}&= \sqrt{ c^2 \left(\mathbf{P} - \frac{e}{c} \mathbf{A}\right) + (m c^2)^2 } + e \phi \\ &= \sqrt{ c^2 (\gamma m \mathbf{v})^2 + (m c^2)^2 } + e \phi \\ &= m c \sqrt{ (\gamma \mathbf{v})^2 + c^2 } + e \phi \\ &= m c \sqrt{ \frac{ \mathbf{v}^2 + c^2 ( 1 - \mathbf{v}^2/c^2) } { 1 - \mathbf{v}^2/c^2 } } + e \phi \\ &= \gamma m c^2 + e \phi \\ &= m c^2 \left( 1 + \frac{1}{{2}} \frac{\mathbf{v}^2}{c^2} + \cdots \right) + e \phi.\end{aligned}

We find that, in the non-relativistic limit, we have

\begin{aligned}\mathcal{E} - e \phi = m c^2 + \frac{1}{{2}} m \mathbf{v}^2 + \cdots \approx m c^2,\end{aligned} \hspace{\stretch{1}}(3.23)

and obtain the first order approximation of our time derivative operator

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\approx\frac{1}{{c^2}} ( \mathcal{E}_{T} - e \phi ) 2 m c^2 u,\end{aligned} \hspace{\stretch{1}}(3.24)

or

\begin{aligned}\frac{1}{{c^2}} ( \mathcal{E} - e \phi )^2 u - (m c)^2 u\approx2 m ( \mathcal{E}_{T} - e \phi ).\end{aligned} \hspace{\stretch{1}}(3.25)

It seems slightly underhanded to use the free particle Hamiltonian in one part of the approximation, and the Hamiltonian for a particle in a field for the other part. This is probably why the text just mandates that $e\phi$ be small compared to $m c^2$.

To summarize once more before the final reduction (where we eliminate the electric field component of the operator equation), we have

\begin{aligned}0=({p} + mc)({p} - mc) u\approx\left(2 m ( \mathcal{E}_{T} - e \phi )- \left(\mathbf{P} - \frac{e}{c} \mathbf{A} \right)^2- \frac{i e \hbar}{c} \boldsymbol{\alpha} \cdot \mathbf{E}+ \frac{e \hbar}{c} \boldsymbol{\sigma}' \cdot \mathbf{H}\right) u.\end{aligned} \hspace{\stretch{1}}(3.26)

Except for the electric field term, this is the result that is derived in the text. It was argued that this term is not significant compared to $e \phi$ when the particle velocity is restricted to the non-relativistic domain. This is done by computing the expectation of this term relative to $e \phi$. Consider

\begin{aligned}{\left\lvert{ \left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \phi } }}\right\rangle }\right\rvert}\end{aligned} \hspace{\stretch{1}}(3.27)

With the velocities low enough so that the time variation of the vector potential does not contribute to the electric field (i.e. the electrostatic case), we have

\begin{aligned}\mathbf{E} = - \boldsymbol{\nabla} \phi = - \hat{\mathbf{r}} \frac{\partial {\phi}}{\partial {r}}.\end{aligned} \hspace{\stretch{1}}(3.28)

The variation in length $a$ that is considered is labeled the characteristic length

\begin{aligned}p a \sim \hbar,\end{aligned} \hspace{\stretch{1}}(3.29)

so that with $p = m v$ we have

\begin{aligned}a \sim \frac{\hbar}{m v}.\end{aligned} \hspace{\stretch{1}}(3.30)

This characteristic length is not elaborated on, but one can observe the similarity to the Compton wavelength

\begin{aligned}L_{\text{Compton}} = \frac{\hbar}{m c},\end{aligned} \hspace{\stretch{1}}(3.31)

the length scale for which Quantum field theory must be considered. This length scale is considerably larger for velocities smaller than the speed of light. For example, the drift velocity of electrons in copper is $\sim 10^{6} \frac{\text{m}}{\text{s}}$, which fixes our length scale to $100$ times the Compton length ($\sim 10^{-12} \text{m})$. This is still a very small length, but is in the QM domain instead of QED. With such a length scale consider the magnitude of a differential contribution to the electric field

\begin{aligned}{\left\lvert{\phi}\right\rvert} = {\left\lvert{\mathbf{E}}\right\rvert} \Delta x = {\left\lvert{\mathbf{E}}\right\rvert} a,\end{aligned} \hspace{\stretch{1}}(3.32)

so that

\begin{aligned}\left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e {\left\lvert{\phi}\right\rvert} } }}\right\rangle&=\left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e a {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle \\ &=\left\langle{{ \frac{e \hbar}{m} \frac{1}{ 2 c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \frac{\hbar }{ m v } {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle \\ &=\frac{1}{{2}} \frac{v}{c} \left\langle{{ \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{ {\left\lvert{\mathbf{E}}\right\rvert} } }}\right\rangle.\end{aligned}

Thus the magnitude of this (vector) expectation is dominated by the expectation of just the $\boldsymbol{\alpha}$. That has been calculated earlier when Dirac currents were considered, where it was found that

\begin{aligned}\left\langle{{\alpha_i}}\right\rangle = \psi^\dagger \alpha_i \psi = (\mathbf{j})_i.\end{aligned} \hspace{\stretch{1}}(3.33)

Also recall that (33.73) that this current was related to momentum with

\begin{aligned}\mathbf{j} = \frac{\mathbf{p}}{m c} = \frac{\mathbf{v}}{c}\end{aligned} \hspace{\stretch{1}}(3.34)

which allows for a final approximation of the magnitude of the electric field term’s expectation value relative to the $e\phi$ term of the Hamiltonian operator. Namely

\begin{aligned}{\left\lvert{ \left\langle{{ \frac{e \hbar}{ 2 m c} \frac{\boldsymbol{\alpha} \cdot \mathbf{E}}{e \phi } }}\right\rangle }\right\rvert}\sim\frac{\mathbf{v}^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(3.35)

With that last approximation made, the gauge transformed Dirac equation, after non-relativistic approximation of the energy and electric field terms, is left as

\begin{aligned}i \hbar \frac{\partial {}}{\partial {t}}=\frac{1}{{2m}} \left(i \hbar \boldsymbol{\nabla} + \frac{e}{c} \mathbf{A} \right)^2- \frac{e \hbar}{2 m c} \boldsymbol{\sigma}' \cdot \mathbf{H}+ e \phi.\end{aligned} \hspace{\stretch{1}}(3.36)

This is still a four dimensional equation, and it is stated in the text that only the large component is relevant (reducing the degrees of spin freedom to two). That argument makes a bit more sense with the matrix form of the gauge reduction which follows in the next section, so understanding that well seems worthwhile, and is the next thing to digest.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

[3] Peeter Joot. Misc Physics and Math Play, chapter Hamiltonian notes. http://sites.google.com/site/peeterjoot/math2009/miscphysics.pdf.

## PHY450H1S. Relativistic Electrodynamics Lecture 8 (Taught by Prof. Erich Poppitz). Relativistic dynamics.

Posted by peeterjoot on February 2, 2011

Covering chapter 2 material from the text [1].

Covering a bit more of Professor Poppitz’s lecture notes: equation of motion, symmetries, and conserved quantities (energy-momentum 4 vector) from relativistic particle action.

Covering lecture notes pp. 56.1-72: comments on mass, energy, momentum, and massless particles (56.1-58); particles in external fields: Lorentz scalar field (59-62); reminder of a vector field under spatial rotations (63) and a Lorentz vector field (64-65) [Tuesday, Feb. 1]; the action for a relativistic particle in an external 4-vector field (65-66); the equation of motion of a relativistic particle in an external electromagnetic (4-vector) field (67,68,73) [Wednesday, Feb. 2]; mathematical interlude: (69-72): on 3×3 antisymmetric matrices, 3-vectors, and totally antisymmetric 3-index tensor – please read by yourselves, preferably by Wed., Feb. 2 class! (this is important, well also soon need the 4-dimensional generalization)

# Finishing previous arguments on action and proper velocity.

For a free particle, our action is

\begin{aligned}S &= - m c \int ds \\ &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}\end{aligned}

Our Lagrangian is

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} .\end{aligned} \hspace{\stretch{1}}(2.1)

We can also make a non-relativistic velocity approximation

\begin{aligned}\mathcal{L} &= -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \\ &= -m c^2 \left( 1 - \frac{1}{{2}} \frac{\mathbf{v}^2}{c^2} \right) + O((\mathbf{v}^2/c^2)^2) \\ &\approx \underbrace{-m c^2 }_{constant} + \underbrace{\frac{1}{{2}} m \mathbf{v}^2 }_{\text{Classical Lagrangian for free particle}}.\end{aligned}

It is good to know that we recover the familiar Newtonian case when our velocities are small enough.

Our job is to vary the action between a pair of spacetime points

\begin{aligned}(t_a, \mathbf{x}_a) \rightarrow (t_b, \mathbf{x}_b)\end{aligned} \hspace{\stretch{1}}(2.2)

The equations of motion that result from this variation, or from the Euler-Lagrange equations that one can obtain from this variation, are

\begin{aligned}\frac{d}{dt} ( \gamma \mathbf{v}) = 0\end{aligned} \hspace{\stretch{1}}(2.3)

We argued last time, by evaluating the derivatives of 2.3, and taking dot and cross products with $\mathbf{v}$ that we also have

\begin{aligned}\frac{d\mathbf{v}}{dt} = 0\end{aligned} \hspace{\stretch{1}}(2.4)

Observe that since $d\mathbf{v}/dt = 0$, we also have $d\gamma/dt = 0$

\begin{aligned}\frac{d\gamma}{dt} &=\frac{d}{dt} \frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \\ &=\frac{d}{dt} \frac{1}{{\left(1 - \frac{\mathbf{v}^2}{c^2}\right)^{3/2}}} (-1/2) (2) (-\mathbf{v} \cdot \dot{\mathbf{v}}) /c^2 \\ &= 0.\end{aligned}

We can therefore combine the pair of equations (after adjusting both to have dimensions of velocity)

\begin{aligned}\frac{d}{dt} ( \gamma \mathbf{v} ) &= 0 \\ \frac{d}{dt} ( \gamma c ) &= 0,\end{aligned} \hspace{\stretch{1}}(2.5)

into

\begin{aligned}u^i = (u^0, \mathbf{u}).\end{aligned} \hspace{\stretch{1}}(2.7)

Here

\begin{aligned}u^0 &= \gamma c \\ \mathbf{u} &= \gamma \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(2.8)

Since we have $du^i/dt = 0$, pre-multiplying this by $\gamma/c$ does not change the equation, and we have

\begin{aligned}0 = \frac{1}{{c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \frac{du^i}{dt}.\end{aligned} \hspace{\stretch{1}}(2.10)

This now puts things in a nice invariant form, with no bias towards any specific observer’s time coordinates, and we have for the free particle

\begin{aligned}\frac{d u^i}{ds} = 0.\end{aligned} \hspace{\stretch{1}}(2.11)

# Symmetries of spacetime translation invariance.

The symmetries of $S$ imply conservation laws. Our action has $SO(1,3) \times T^4 =$ Lorentz x spacetime translation $\equiv$ Poincar\'{e} group of symmetries.

Consider quantities conserved due to $T^4$ factor

\begin{aligned}\mathbf{x} &\rightarrow \mathbf{x} + \mathbf{a} \qquad \mbox {wherelatex \mathbf{a}is constant} \\ t &\rightarrow t + \text{constant}\end{aligned} \hspace{\stretch{1}}(3.12)

Observe that the Lagrangian is not a function of $\mathbf{x}$, or $t$ explicitly

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = - m c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}= \mathcal{L}(\mathbf{v}) .\end{aligned} \hspace{\stretch{1}}(3.14)

A consequence from this, utilizing the Euler-Lagrange equations is that we have a zero for the time derivative of the generalized momentum ${\partial {\mathcal{L}}}/{\partial {\mathbf{v}}}$

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = 0,\end{aligned} \hspace{\stretch{1}}(3.15)

Let’s calculate that generalized momentum

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} &=\frac{\partial {}}{\partial {\mathbf{v}}} \left(-m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \right) \\ &=\frac{\partial {}}{\partial {\mathbf{v}}} \left( -m c^2 \frac{(1/2)(-2) \mathbf{v}/c^2}{ \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} } \right) \\ &=m \frac{\mathbf{v}}{ \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} } \\ \end{aligned}

So our generalized momentum is

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = m \mathbf{v} \gamma.\end{aligned} \hspace{\stretch{1}}(3.16)

Evaluating the Euler-Lagrange equations above we find

\begin{aligned}0 &= \frac{d}{dt} \left( m \gamma \mathbf{v} \right) \\ &= \frac{d}{dt} \left( m u^{1,2,3} \right) \\ \end{aligned}

Recall that $u^0 = c \gamma$, and that $d\gamma/dt = 0$, so we also have

\begin{aligned}\frac{d}{dt} \left( m u^i \right) = 0\end{aligned} \hspace{\stretch{1}}(3.17)

and again with multiplication by $\gamma/c$ we have a Lorentz invariant relation, mostly a consequence of spacetime translation invariance

\begin{aligned}\frac{d}{ds} \left( m u^i \right) = 0.\end{aligned} \hspace{\stretch{1}}(3.18)

We define this quantity, the invariant quantity (a four vector), as the relativistic momentum

\begin{aligned}p^i = m u^i.\end{aligned} \hspace{\stretch{1}}(3.19)

A relativistic particle is characterizes by a conserved 4 vector quantity $p^i$ with

\begin{aligned}p^0 &= m c \gamma \\ \mathbf{p} &= m \gamma \mathbf{v} \\ p^i &= (p^0, \mathbf{p})\end{aligned} \hspace{\stretch{1}}(3.20)

# Time translation invariance

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = \mathcal{L}(\mathbf{v}) \end{aligned} \hspace{\stretch{1}}(4.23)

However, it helps to consider the more general case

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = \mathcal{L}(\mathbf{x}, \mathbf{v}) \end{aligned} \hspace{\stretch{1}}(4.24)

since we have no explicit time dependence.

\begin{aligned}\frac{d}{dt} \mathcal{L}(\mathbf{v}) &= \frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} \cdot \dot{\mathbf{x}} + \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \dot{\mathbf{v}} \\ &= \left( \frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \right) \cdot \mathbf{v} + \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \frac{d\mathbf{v}}{dt} \\ &= \frac{d}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathbf{v}\right)\end{aligned}

Regrouping, to pull all the derivative terms together provides the conservation identity

\begin{aligned}\frac{d}{dt} \left(\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathbf{v} - \mathcal{L}\right) = 0.\end{aligned} \hspace{\stretch{1}}(4.25)

This quantity $\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathbf{v} - \mathcal{L}$ is usually identified as the Hamiltonian $H$, the energy, but we will call it $E$ here.

In our case, with the relativistic free particle Lagrangian

\begin{aligned}\mathcal{L} = -m c^2 \sqrt{ 1 - \frac{\mathbf{v}^2}{c^2} },\end{aligned} \hspace{\stretch{1}}(4.26)

we have

\begin{aligned}E &= \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathbf{v} - \mathcal{L} \\ &= \mathbf{v} \cdot \left( m \frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \mathbf{v} \right) + m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \\ &= \frac{m \mathbf{v}^2}{\sqrt{ 1 - \frac{\mathbf{v}^2}{c^2}}} + mc^2 \sqrt{ 1 - \frac{\mathbf{v}^2}{c^2}} \\ &= \frac{\mathbf{v}^2 + m c^2 \left(1 - \frac{\mathbf{v}^2}{c^2} \right)}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \\ &= \frac{m c^2 }{\sqrt{ 1 - \frac{\mathbf{v}^2}{c^2} }}\end{aligned}

So we define, for the energy, a conserved quantity under time translation, we have

\begin{aligned}\boxed{E = \gamma m c^2 = \frac{ m c^2 }{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}}\end{aligned} \hspace{\stretch{1}}(4.27)

It is only with the $\mathbf{v} \rightarrow 0$ that we recover the famous tee-shirt expression

\begin{aligned}E = m c^2.\end{aligned} \hspace{\stretch{1}}(4.28)

Since we also know (from the spacetime translation) that $p^0 = m c \gamma = E/c$, we get another conserved quantity for free since $(p^0, \mathbf{p})$ then is also a symmetry (i.e. thus a conserved quantity)

\begin{aligned}p^0 &= m \gamma c = \frac{E}{c} \\ \mathbf{p} &= m \gamma \mathbf{v}\end{aligned}

\begin{aligned}p^i = ( p^0, \mathbf{p} )\end{aligned} \hspace{\stretch{1}}(4.29)

Note that the only “mass” you ever want to talk about is “m”. This is a Lorentz scalar, and we won’t use the old notions that mass changes with velocity or “relativistic mass”.

# Some properties of the four momentum.

We have

\begin{aligned}p^i p_i &= (p^0)^2 - \mathbf{p}^2 \\ &= m c^2 \gamma^2 - m^2 \gamma^2 \mathbf{v}^2 \\ &= m c^2 \gamma^2 \left( 1 - \frac{\mathbf{v}^2}{c^2} \right) \\ &= m^2 c^2\end{aligned}

So we have

\begin{aligned}\boxed{p^i p_i = m^2 c^2}\end{aligned} \hspace{\stretch{1}}(5.30)

We say that the 4-vector $p^i$ represents a particle with mass $m$.

Since four momentum is a conserved quantity we can use this conservation property to study relativistic collisions

PICTURE: two particles colliding with two particles resulting (particles trajectories as arrows)

\begin{aligned}\underbrace{p_1^i + p_2^i}_{\text{four momentum before}} = \underbrace{p_3^i + p_4^i}_{\text{four momentum after}}\end{aligned} \hspace{\stretch{1}}(5.31)

\begin{aligned}\mathbf{p} &= \frac{m \mathbf{v}}{\sqrt{ 1 - \frac{\mathbf{v}^2}{c^2} }} \rightarrow 0 \qquad \mbox{whenlatex m \rightarrow 0} \\ E &= \frac{m c^2}{\sqrt{ 1 – \frac{\mathbf{v}^2}{c^2} }} \rightarrow 0 \qquad \mbox{when $m \rightarrow 0$}\end{aligned} \hspace{\stretch{1}}(5.32)

except when ${\left\lvert{\mathbf{v}}\right\rvert} = c$, where if you take $m \rightarrow 0$ and ${\left\lvert{\mathbf{v}}\right\rvert} = c$ you can get anything (any values) in such a limit (limit does not exist).

However, because

\begin{aligned}\frac{E^2}{c^2} - \mathbf{p}^2 = m^2 c^2 = 0\end{aligned} \hspace{\stretch{1}}(5.34)

when $m \rightarrow 0$, $E$ and $\mathbf{p}$ for a massless particle must obey $E = c {\left\lvert{\mathbf{p}}\right\rvert}$.

Massless particles like photons (and gravitons if/when eventually measured) have lightlike 4 momentum vectors

\begin{aligned}p^i p_i = 0\end{aligned} \hspace{\stretch{1}}(5.35)

Gravity waves haven’t been seen yet, but the LIGO and LISA (extremely large infraferometers) experiments are expected to get some results on this in the near future.

# Where are we?

In the notes there’s a review (see that on one’s own). We’ll also want to eventually deal with the conservation laws in four vector form, since it will illustrate how the electric and magnetic fields have to be transformed. We’ll get to that eventually.

# Interactions

In classical mechanics we have

\begin{aligned}\mathcal{L}_\text{kinetic} = \frac{1}{{2}} m \mathbf{v}^2 \end{aligned} \hspace{\stretch{1}}(7.36)

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \mathbf{v}^2 - U(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(7.37)

Here $U(\mathbf{r})$ is an external potential.

\begin{aligned}S = S_{\text{free}} + S_{\text{interaction}} = \int dt \frac{1}{{2}} m \mathbf{v}^2 + \int dt (-U(\mathbf{r}, t))\end{aligned} \hspace{\stretch{1}}(7.38)

The quantity $U(\mathbf{r}, t)$ is what we call a potential field.

What’s the simplest invariant field we can have? The simplest possibility is to have a relativistic particle which interacts with an external \underline{Lorentz scalar field}. We’d imagine that this is due to some other particle or some distribution of other fields.

Recall that the scalar field under rotations (reminder)

PICTURE: a point with coordinates in a fixed and a rotated coordinate system

That point is

\begin{aligned}P = (x, y) = (x', y')\end{aligned} \hspace{\stretch{1}}(7.39)

Similarly we can define a scalar quantity (like temperature or the Coulomb potential) is then assigned a value at each point

\begin{aligned}\phi(x, y) = \phi'(x', y')\end{aligned} \hspace{\stretch{1}}(7.40)

The value of this scalar in the $x,y$ coordinates system at point $P$ equals the value of this scalar in the $x',y'$ coordinates system at the same point $P$.

A Lorentz scalar field is like this, but for an event $P = (ct, x) = (ct', x')$ is the same.

So, we’d have

\begin{aligned}\phi(ct, x) = \phi'(ct', x')\end{aligned} \hspace{\stretch{1}}(7.41)

The value of this scalar in the $x,ct$ coordinates system at event $P$ equals the value of this scalar in the $x',ct'$ coordinates system at the same event $P$ in the primed frame.

Our action would then be

\begin{aligned}S = - m c \int ds + g \int ds \phi(x^i)\end{aligned} \hspace{\stretch{1}}(7.42)

Here $g$ is a coupling constant, also called the “charge” of a particle under that scalar field.

Note that unfortunately nature hasn’t provided us with scalar fields that are stable enough to observe in classical interactions

We do however have some scalar particles

\begin{aligned}\pi^0, \pi^\pm, k^0, k^\pm\end{aligned} \hspace{\stretch{1}}(7.43)

These are unstable and short ranged.

The LHC is looking for another unstable short lived scalar field (the Higgs). So we have to unfortunately study a more complicated field, a vector field. We’ll do that next time.

# References

[1] L.D. Landau and E.M. Lifshits. The classical theory of fields. Butterworth-Heinemann, 1980.